Proposition 21

Suppose that $u_N$ is of finite trace. Then there exist a set $\widetilde{X}\subset X$ whose complement X $-\widetilde{X}$ is $\mu$-negligible and a function $H\in \mathscr{L}^2(X\times X)$ satisfying the following conditions :

(i) For every $(x, y)\in \widetilde{X}\times \widetilde{X}$, the family $(\alpha_if_i(x)g_i(y))_{i\in I}$ is summable in $\mathbf{C}$ and its sum is $N(x, y)$;

(ii) For every finite subset J of I and every $(x, y)\in \widetilde{X}\times \widetilde{X}$, one has

(iii) The function $x\mapsto H(x, x)$ belongs to $\mathscr{L}^1(X)$.

Since $u_N$ is of finite trace, the family $(\alpha_i)$ is summable (cor. 4 of IV, p. 165). The series

therefore converge in $\mathscr{L}^1(X)$ and consequently $\mu$-almost everywhere (INT, IV, p. 128, § 3, n$^o3$, prop. 6). Let F and G denote, respectively, the functions defined $\mu$-almost everywhere by the sum of these series, putting $F(x) = 0$ and $G(x) = 0$ for every $x$ such that the corresponding series does not converge. Since F and G belong to $\mathscr{L}^1$(X), the function H defined by $H(x, y) =\surd F(x)G(y)$ belongs to $\mathscr{L}^2(X\times X)$ (INT, V, p. 95, § 8, n$^o3$, cor. 2).

Let us apply prop. 20 to the space X, to $F =\mathbf{C}^2$, and to the measurable mappings

$i\leqslant ni\leqslant n$

defined on X. There therefore exists a sequence $(C_m)_{m\in\mathbf{N}}$ of compact subsets of X satisfying the following conditions:

(1) the union $\widetilde{X}$ satisfies $\mu(X-\widetilde{X}) = 0$;

(2) for every $m\in \mathbf{N}$ and every $n\in \mathbf{N}$, the functions $s_n$ are continuous on $C_m$;

(3) for every $m\in \mathbf{N}$, the series $\sum\alpha_i|f_i|^2$ and $\sum\alpha_i|g_i|^2$ converge uniformly on $C_m$ to F and G, respectively;

(4) the support of the measure induced by $\mu$ on $C_m$ is equal to $C_m$ (replacing $C_m$ by the support of this measure if necessary).

Let us prove that the set $\widetilde{X}$ and the function H satisfy conditions (i), (ii), and (iii).

Let $(x, y)\in \widetilde{X}\times \widetilde{X}$. For every finite subset $J\subset I$, one has

whence also

which already establishes property (ii).

By inequality (15), the series $\sum_i\alpha_ih_i$ converges uniformly on $K_m\times K_n$ for every $(m, n)\in \mathbf{N}^2$. Let $\widetilde{N}$ be the function on $X\times X$ defined by

for every $(x, y)\in \widetilde{X}\times \widetilde{X}$ and by $\widetilde{N}(x, y) = 0$ otherwise. The function $\widetilde{N}$ is measurable (INT, IV, p. 175, § 5, n$^o4$, th. 2), and it is continuous on $K_m\times K_n$ for every $(m, n)\in \mathbf{N}^2$.

From (16), we have $|\widetilde{N}(x, y)|\leqslant H(x, y)$ for all $(x, y)\in X\times X$, and in particular $\widetilde{N}$ belongs to $\mathscr{L}^2(X\times X)$ (INT, IV, p. 84, § 5, n$^o6$, th. 5).

Let us prove that $N =\widetilde{N}$ on $\widetilde{X}\times \widetilde{X}$, which will establish property (i). Let $f$ and $g$ be elements of $\mathscr{L}^2(X)$. We have

(formula (12) of IV, p. 172). For all $(x, y)\in \widetilde{X}\times \widetilde{X}$ and every finite subset J of I, we have

by formula (16). Since the right-hand side of this inequality is integrable on $X\times X$ (INT, V, p. 95, § 8, n$^o3$, cor. 2), we may apply Lebesgue's theorem (INT, IV, p. 137, § 3, n$^o7$, th. 6) and formula (13), and deduce that

Consequently, we obtain $u_{\widetilde{N}}=u_N$, whence $N =\widetilde{N}$ in $L^2(X\times X)$ (prop. 3 of III, p. 28, b)).

For all $(m, n)\in \mathbf{N}^2$, the functions N and $\widetilde{N}$ are continuous on $K_m\times K_n$ and therefore are equal on $K_m\times K_n$ (prop. 9 of INT, III, p. 69, § 2, n$^o2$), since the support of the measure induced by $\mu\otimes \mu$ on $K_m\times K_n$ is equal to $K_m\times K_n$. Hence N coincides with $\widetilde{N}$ on $\widetilde{X}\times \widetilde{X}$.

Finally, the estimate FG $\leqslant (F + G)/2$ implies that the function $x\mapsto \surd F(x)G(x) = H(x, x)$ is integrable on X, whence property (iii).

In the rest of this number, we retain the notation of the proposition.